求当前结点到根节点的距离 Sample Input 2 1 //n m 1 2 1 2 //询问 5 2 1 2 1 3 3 4 3 5 4 2 //询问 4 5 0 0 Sample Output lxh pfz lxh # include <iostream> # include <cstdio> # include <cstring> # include <algorithm> # include <cmath> # include <…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2545 思路:dist[u]表示节点u到根节点的距离,然后在查找的时候更新即可,最后判断dist[u],dist[v]的大小即可. #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<vector> using namespace std;…

A very big corporation is developing its corporative network. In the beginning each of the N enterprises of thecorporation, numerated from 1 to N, organized its own computing and telecommunication center. Soon, foramelioration of the services, the co…

根节点的Level属性为0,一级子节点Level属性为1,二级子节点Level属性为2,以此类推:同级节点可以用索引.名称.文本来区分.用索引区分根节点时,TreeView.Nodes[0]就是第一个根节点,TreeView.Nodes[1]就是第二个根节点,以此类推:用索引区分一级子节点时,TreeView.Nodes[0].Nodes[0]为第一个根节点的第一个子节点,TreeView.Nodes[0].Nodes[1]是第一个根节点的第二个子节点,以此类推: 2.如何获取TreeView点…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1890 Time Limit: 6000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Problem Description Somewhere deep in the Czech Technical University buildings, there are laboratories for examining…

首先定义一个节点类,包含三个成员变量,分别是节点值,左指针,右指针,如下代码所示: class Node(object): def __init__(self, value): self.value = value self.left = None self.right = None 接下来就是二叉树的相关工作: 1)初始化一棵二叉树 class Solution(object): def __init__(self): root = Node(6) B = Node(4) root.left…

如果二叉树每个结点的权值关于根节点完全对称 就输出Yes Sample Input 27 //结点1 2 3 //结点1的左孩子是结点2 右孩子是结点32 4 53 6 74 0 05 0 06 0 07 0 01 2 2 3 4 4 3 //权值51 2 32 0 43 0 54 0 05 0 01 2 2 3 3 Sample Output YesNo # include <cstdio> # include <cstring> # define LL long long us…

原创博文,转载请注明出处! # 题目 # 举例        下图二叉树的深度为4,最长路径为1-2-5-7. # 思路(递归)       如果一个树只有一个节点,它的深度为1: 如果根节点只有左子树而没有右子树,那么树的深度应该是其左子树的深度+1: 如果根节点只有右子树而没有左子树,那么树的深度应该是其右子树的深度+1: 如果根节点既有左子树又有右子树,那么树的深度应该是左右字数深度的较大值+1. # 代码 class Solution { public: int TreeDepth(Tr…

原创博文,转载请注明出处! # 题目 # 举例 # 思路 由平衡二叉树的定义可知,判断二叉树是否是平衡二叉树的关键在于判断任意结点是否是平衡结点.后序遍历二叉树,判断节点的子树是否平衡并计算节点的子树高度,判断结点是否平衡.如果按后序遍历的顺序,遍历到根节点后,根节点也是平衡结点,则二叉树是平衡二叉树.注意:叶子节点是平衡节点,叶子结点高度为1. # 代码 //后续遍历二叉树,遍历过程中求子树高度,判断是否平衡 class Solution { public: bool IsBalanced(T…

Problem Description Acesrc is a famous tourist at Nanjing University second to none. During this summer holiday, he, along with Zhang and Liu, is going to travel to Hong Kong. There are n spots in Hong Kong, and n−1 bidirectional sightseeing bus rout…